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The Henderson-Hasselbalch equation

Once you know a buffer is present, its pH is one line of arithmetic: pKa plus the log of the base-to-acid ratio. Get the ratio right-side-up and the pH falls out.

§1

pH from the component ratio.

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The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), gives a buffer's pH from the ratio of conjugate base [A⁻] to weak acid [HA].

When the ratio is 1 (equal amounts), the log is 0 and pH = pKa. More conjugate base than acid makes the log positive, raising pH above pKa; more acid makes it negative, lowering pH below pKa.

The conjugate base [A⁻] is the numerator — getting the ratio the right way up is the crux of applying the equation.

UNIT 8 TOPIC 8.9 • HENDERSON-HASSELBALCH EQUATION HENDERSON-HASSELBALCH The acid-base ratio [A⁻]/[HA] shifts pH around pKa. pH = pKa + log([A⁻]/[HA]) where pKa = − log Ka [A⁻]/[HA] = 0.10 log 0.1 = −1 HA A⁻ more HA (acid) pH = pKa − 1 [A⁻]/[HA] = 1 log 1 = 0 HA A⁻ equal amounts pH = pKa [A⁻]/[HA] = 10 log 10 = +1 HA A⁻ more A⁻ (base) pH = pKa + 1 CED ANCHOR Valid only for buffers — an appreciable amount of BOTH the weak acid (HA) and its conjugate base (A⁻) must be present. AP Chemistry · Unit 8 · Acids and Bases
Fig. 8.9.1 The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), gives a buffer's pH from the ratio of conjugate base to weak acid. When the ratio is 1, pH = pKa; more conjugate base raises the pH, more acid lowers it.
§2

Using the equation.

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Put the conjugate base on top, then take the log.

  1. Identify [A⁻] and [HA]. The conjugate base and the weak acid concentrations.
  2. Form the ratio [A⁻]/[HA]. Conjugate base over weak acid (base on top).
  3. Take the log and add pKa. pH = pKa + log([A⁻]/[HA]).
  4. Sanity-check. Ratio 1 → pH = pKa; more base → above pKa; more acid → below pKa.
§3

The pieces you'll meet.

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One equation, one correct ratio.

H-H
Henderson-Hasselbalch
pH = pKa + log([A⁻]/[HA]).
A⁻
[A⁻]
Conjugate base — the numerator.
HA
[HA]
Weak acid — the denominator.
ratio 1
Ratio = 1
pH = pKa (log 0).
more base
More base
Raises pH above pKa.
more acid
More acid
Lowers pH below pKa.
§4

Worked example: buffer pH.

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Given. A buffer has pKa = 4.7, with [A⁻] = 0.20 M and [HA] = 0.20 M.

Ratio. [A⁻]/[HA] = 0.20/0.20 = 1, so log(1) = 0.

pH. pH = 4.7 + 0 = 4.7 = pKa, since the components are equal.

Shift. If instead [A⁻]/[HA] = 10, then log(10) = 1 and pH = 4.7 + 1 = 5.7 — higher, because there is more conjugate base.

§5

Mistakes that cost real points.

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Pitfall · 01

"Put the weak acid on top: pH = pKa + log([HA]/[A⁻])."

The conjugate base [A⁻] goes in the numerator: pH = pKa + log([A⁻]/[HA]). Inverting the ratio flips the sign of the log, giving a pH shifted the wrong way from pKa.

Fix. Keep the conjugate base on top: pH = pKa + log([A⁻]/[HA]).

Pitfall · 02

"Henderson-Hasselbalch applies to any solution."

The equation applies to buffers — solutions with significant amounts of both a weak acid and its conjugate base. It is not valid for a strong acid, a lone weak acid, or far outside the buffer region.

Fix. Use Henderson-Hasselbalch only for buffers with both components present.

Pitfall · 03

"When [A⁻] = [HA], the pH is 7."

When the components are equal, log(1) = 0 and pH = pKa — not necessarily 7. A buffer with pKa = 4.7 has pH 4.7 when the components are equal, which is acidic.

Fix. Read equal components as pH = pKa, whatever that pKa value is, not pH 7.

§6

Skill Check.

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